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# Ask a Mathematics(Class XI-XII Tuition (PUC)) Question

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Vishal 14/11/2016 in  Mathematics(Class XI-XII Tuition (PUC)),

What is latus rectum?

Sarvajeet replied | 26/11/2016

In any conics (i.e., parabola or ellipse or hyperbola ), a line passing through its focus and parallel to its directrix is called the latus rectum.
For example, in parabola: square of y = 4ax, the equation of the latus ractum y=a and its length = 4a;
in ellipse: square of (x/a) + square of (y/b)=1, a>b, the equation of the latus ractum y=ae, -ae and its length = 2(square...  more»
In any conics (i.e., parabola or ellipse or hyperbola ), a line passing through its focus and parallel to its directrix is called the latus rectum.
For example, in parabola: square of y = 4ax, the equation of the latus ractum y=a and its length = 4a;
in ellipse: square of (x/a) + square of (y/b)=1, a>b, the equation of the latus ractum y=ae, -ae and its length = 2(square of b)/a;
in hyperbola: square of (x/a) - square of (y/b)=1, the equation of the latus ractum y=ae, -ae and its length = 2(square of b)/a; «less

Ankit replied | 09/12/2016

In any conics (i.e., parabola or ellipse or hyperbola ), a line passing through its focus and parallel to its directrix is called the latus rectum.
For example, in parabola: square of y = 4ax, the equation of the latus ractum y=a and its length = 4a;
in ellipse: square of (x/a) + square of (y/b)=1, a>b, the equation of the latus ractum y=ae, -ae and its length = 2(square...  more»
In any conics (i.e., parabola or ellipse or hyperbola ), a line passing through its focus and parallel to its directrix is called the latus rectum.
For example, in parabola: square of y = 4ax, the equation of the latus ractum y=a and its length = 4a;
in ellipse: square of (x/a) + square of (y/b)=1, a>b, the equation of the latus ractum y=ae, -ae and its length = 2(square of b)/a;
in hyperbola: square of (x/a) - square of (y/b)=1, the equation of the latus ractum y=ae, -ae and its length = 2(square of b)/a; «less

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U

Uday 14/11/2016 in  Mathematics(Class XI-XII Tuition (PUC)),

What are the terms related to parabola?

Abhilash replied | 02/12/2016

u-shaped curve with certain specific properties. Formally, a parabola is defined as follows:- For a given point, called the focus, and a given line not through the focus, called the directrix, a parabola is the locus of points such that the distance to the focus equals the distance to the directrix.

Ankit replied | 09/12/2016

u-shaped curve with certain specific properties. Formally, a parabola is defined as follows:- For a given point, called the focus, and a given line not through the focus, called the directrix, a parabola is the locus of points such that the distance to the focus equals the distance to the directrix.

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J

Jogendra 14/11/2016 in  Mathematics(Class XI-XII Tuition (PUC)),

What is the parametric equation of a chord?

Vanaja replied | 29/11/2016

The chord of contact is a chord formed from joining the points of intersection of the parabola and two tangents emanating from an external point T. Both methods are used to derive the equation of the chord of contact.

Ankit replied | 09/12/2016

The chord of contact is a chord formed from joining the points of intersection of the parabola and two tangents emanating from an external point T. Both methods are used to derive the equation of the chord of contact.

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T

Tashiruddin 14/11/2016 in  Mathematics(Class XI-XII Tuition (PUC)),

What do you mean by curve with respect to parabola?

Abhilash replied | 01/12/2016

A parabola is a curve where any point is at an equal distance from: a fixed point (the focus ) and a fixed straight line (the directrix).

Ankit replied | 09/12/2016

A parabola is a curve where any point is at an equal distance from: a fixed point (the focus ) and a fixed straight line (the directrix).

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S

Shreyansh 14/11/2016 in  Mathematics(Class XI-XII Tuition (PUC)),

What do you mean by conormal point?

Prateek replied | 19/11/2016

The points on the curve at which the normal pass through a common point are called co-normal points.

Sanjeev replied | 12/12/2016

Perpendicular to each other.

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A

Ambar 14/11/2016 in  Mathematics(Class XI-XII Tuition (PUC)),

Find the vertex, axis, focus, directrix, latus rectum of the parabola: (square of y) + 8y + x + 19 = 0?

Sarvajeet replied | 26/11/2016

It's vertex = (-3, -4),
axis y=-4,
focus (-13/4, -4),
Directrix x= -11/4,
the equation of the Latus rectum x=-13/4 and the length of the Latus rectum=1.

Ankit replied | 09/12/2016

It's vertex = (-3, -4),
axis y=-4,
focus (-13/4, -4),
Directrix x= -11/4,
the equation of the Latus rectum x=-13/4 and the length of the Latus rectum=1.

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R

Rayaan 14/11/2016 in  Mathematics(Class XI-XII Tuition (PUC)),

Find the equation of the chord of a parabola with respect to the point (x, y)?

K.v.ratnamala replied | 07/12/2016

y*y=4ax.

Rajneesh replied | 13/12/2016

y*y = 4ax.

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K

Khushboo 14/11/2016 in  Mathematics(Class XI-XII Tuition (PUC)),

Find the equation of tangents to the parabola drawn from an external point P?

K.v.ratnamala replied | 07/12/2016

if (x,y) is the point then y*y - 4ax is the tangent to the parabola.

Manibhushan replied | 07/12/2016

The equation of the parabola is x^2=4ay or y=x^2/(4a) . We are drawing tangent lines through the point (2,-3) to the parabola.
The slopes of the tangent lines are given by the first derivative evaluated at the point of tangency.
y=x^2/(4a)==>y'=x/(2a).
Thus the equation of a tangent line to the parabola at (x,x^2/(4a)) and through point (2,-3) is y+3=x/(2a)(x-2) or y=x^2/(2a)-x/a-3....  more»
The equation of the parabola is x^2=4ay or y=x^2/(4a) . We are drawing tangent lines through the point (2,-3) to the parabola.
The slopes of the tangent lines are given by the first derivative evaluated at the point of tangency.
y=x^2/(4a)==>y'=x/(2a).
Thus the equation of a tangent line to the parabola at (x,x^2/(4a)) and through point (2,-3) is y+3=x/(2a)(x-2) or y=x^2/(2a)-x/a-3. But y=(x^2)/(4a). So,
x^2/(4a)=x^2/(2a)-x/a-3 Multiplying through by 4a we get:
x^2=2x^2-4x-12a or x^2-4x-12a=0 . Using the quadratic formula we get:
x=(4+-sqrt(16-4(1)(-12a)))/(2)
x=(4+-sqrt(16(1+3a)))/2
x=2+-2sqrt(1+3a). «less

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A

Abishiek 14/11/2016 in  Mathematics(Class XI-XII Tuition (PUC)),

Find the equation of the chord of parabola whose midpoint is (x1, y1)?

Sarvajeet replied | 01/12/2016

The equation of the chord of parabola (Let square of y=4ax) whose midpoint is (x1, y1) is:-
y(y1)-4a(x1)= square of (y1)-4a(x1).

Sarvajeet replied | 03/12/2016

The equation of the chord of parabola (Let square of y=4ax) whose midpoint is (x1, y1) is:-
y(y1)-2a(x-x1)= square of (y1)-4a(x1).

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H

Himangi 14/11/2016 in  Mathematics(Class XI-XII Tuition (PUC)),

Define the horizontal ellipse?

S

Srinivas replied | 28/11/2016

In ellipse major axis is x-axis major axis length = 2a
minor axis is y- axis its length = 2b.

Sarvajeet replied | 03/12/2016

square of (x/a) + square of (y/b)=1, a>b.

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R

Rema 14/11/2016 in  Mathematics(Class XI-XII Tuition (PUC)),

What is the special form of ellipse having direction of the axes are parallel to the coordinate axes?

S

Sudarshan Menon replied | 03/12/2016

circle.

Sarvajeet replied | 03/12/2016

The special form of an ellipse: square of (x/a)+square of (y/b)=1, a>b;
The major axis is (2a) on x-axis and minor axis (2b) y-axis.

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R

Rekha 14/11/2016 in  Mathematics(Class XI-XII Tuition (PUC)),

What is the equation of the chord with respect to ellipse?

Vanaja replied | 29/11/2016

An ellipse is the locus of all points of the plane whose distances to two fixed points add to the same constant.

An ellipse can be obtained as a result from the intersection of a cone by a plane in a way that produces a closed curve. Circles are special cases of ellipses, obtained when the cutting plane is orthogonal to the cone's axis.
If equation of an ellipse...  more»
An ellipse is the locus of all points of the plane whose distances to two fixed points add to the same constant.

An ellipse can be obtained as a result from the intersection of a cone by a plane in a way that produces a closed curve. Circles are special cases of ellipses, obtained when the cutting plane is orthogonal to the cone's axis.
If equation of an ellipse is x2 / a2 + y2 / b2 = 1, then equation of director circle is x2 + y2 = a2 + b2. «less

T

Tamil replied | 29/11/2016

X^2+y^2=a^2+b^2.

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K

Krishna 14/11/2016 in  Mathematics(Class XI-XII Tuition (PUC)),

What is the difference between the diameter and conjugate diameter with respect to ellipse?

Manibhushan replied | 07/12/2016

Any chord that passes through the center of an ellipse is call its diameter. It follows that the family of parallel chords define two diameters: one in the direction to which they are all parallel and the other the locus of their midpoints. Such two diameters are called conjugate.The statement is easily proved analytically if we start with the equation x /a + y /b = 1 and the...  more»
Any chord that passes through the center of an ellipse is call its diameter. It follows that the family of parallel chords define two diameters: one in the direction to which they are all parallel and the other the locus of their midpoints. Such two diameters are called conjugate.The statement is easily proved analytically if we start with the equation x²/a² + y²/b² = 1 and the associated parameterization: x = a cos(t), y = b sin(t).
Thus ellipse is a curve defined by the radius-vector:-
r(t) = (a cos(t), b sin(t)).
For a fixed t, we are interested in two points, r(t ± v). We shall use the addition formulas for sine and cosine:
r(t + v) = (a (cos(t)cos(v) - sin(t)sin(v)), b (sin(t)cos(v) + cos(t)sin(v))),
r(t - v) = (a (cos(t)cos(v) + sin(t)sin(v)), b (sin(t)cos(v) - cos(t)sin(v))).
The slope of the difference, say, r(t + v) - r(t - v) is -b/a cot(t), independent of v, meaning that we thus produce a family of parallel chords. Their midpoints satisfy:-
(r(t + v) + r(t - v)) / 2 = cos(v) (a cos(t), b sin(t))
which is a parameterization (with parameter cos(v)) of the chord with the slope of b/a tan(t). To summarize, the midpoints of the chords parallel to the direction with the slope -b/a cot(t) lie on the line with the slope b/a tan(t). Applying the formulas of sine and cosine of the complementary angles we see that starting with the chords with the latter slope we would have found their midpoints on a line with the former slope, thus justifying the symmetric terminology. The two directions are conjugate. Observe in passing that, although the conjugate diameters correspond to the complementary values of the parameter t, the product of the two slopes is -b²/a² which is -1 only for circles, i.e. when a = b, so that in general, the conjugate diameters are not perpendicular.
Making use of the parameterization r(t) = (a cos(t), b sin(t)) I tacitly assumed that the origin of the system of coordinates has been placed at the center of the ellipse. We now evaluate the distance from the center to the end points of the conjugate diameters, i.e., P = (a cos(t), b sin(t)) and, say, Q = (-a sin(t), b cos(t)):
OP² + OQ² = (a² cos²(t) + b² sin²(t)) + (a² sin²(t) + b² cos²(t))
= (a² cos²(t) + a² sin²(t)) + (b² sin²(t) + b² cos²(t))
= a² + b²,
independent of t. This is known as the first theorem of Apollonius: For the conjugate (semi)diameters OP and OQ, OP² + OQ² = a² + b². «less

Mayank replied | 07/12/2016

Consider family of chords which is parallel to the diameter (chord passing through the center of the ellipse) of the ellipse. Joining the mid points of these chords the line thus formed is the conjugate diameter of the ellipse.

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A

Ashish 14/11/2016 in  Mathematics(Class XI-XII Tuition (PUC)),

What are the properties of conjugate diameters?

Vanaja replied | 29/11/2016

Any chord that passes through the center of an ellipse is call its diameter. It follows that the family of parallel chords define two diameters: One in the direction to which they are all parallel and the other the locus of their midpoints. Such two diameters are called conjugate.

In geometry, two diameters of a conic section are said to be conjugate if each chord parallel to one diameter is bisected by the other diameter. For example, two diameters of a circle are conjugate if and only if they are perpendicular.

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A

Aswini 14/11/2016 in  Mathematics(Class XI-XII Tuition (PUC)),

Length of major axis 26, foci (5, 0) find the equation for the ellipse that satisfies the given condition?

S

Shiv replied | 03/12/2016

length of major axis 2a=26;
a=13;
f=5;
(square of f)=(square of a)-(square of b);
25=169-(square of b);
(square of b)=169-25;
b=12;
the (sauare of x)/(169)-(square of y)/(144)=1

Sarvajeet replied | 03/12/2016

The length of major axis 2a=26;
a=13;
ae=5;
So, e=5/13
square of b=(square of a)(1-(square of e))
So, b=12;
square of (x/13)-square of (y/12)=1.
(square of f)=(square of a)-(square of b);
25=169-(square of b);
(square of b)=169-25;
b=12;
the (square of x)/(169)-(square of y)/(144)=1.

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P

Prasenjit 14/11/2016 in  Mathematics(Class XI-XII Tuition (PUC)),

Find the eccentricity and the length of the latus rectum of the ellipse: 16(square of x) + (square of y) = 16?

Barsha replied | 18/11/2016

Sarvajeet replied | 03/12/2016

The eccentricity = (square root 15)/4;
The length of the latus rectum of the ellipse = 1/2

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D

Dr 14/11/2016 in  Mathematics(Class XI-XII Tuition (PUC)),

Find the major axis on the x - axis and phases through the points (4, 3) and (6, 2)?

S

Shiv replied | 19/11/2016

x+2y=10;
on x-axis is 10.
on y-axis is 5 and angle is tan[inverse(1/2)].

Ankit replied | 09/12/2016

x+2y=10;
on x-axis is 10.
on y-axis is 5 and angle is tan[inverse(1/2)].

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K

Kiran 14/11/2016 in  Mathematics(Class XI-XII Tuition (PUC)),

Find the foci on a x - axis if b = 3, c = 4?

Sagar replied | 21/11/2016

Eti replied | 21/11/2016

Foci is (4,0) and (-4,0).

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N

Neha 14/11/2016 in  Mathematics(Class XI-XII Tuition (PUC)),

Find the locus of the extremities of the latus recta of all ellipses having a given major axis 6a?

Saurabh replied | 01/12/2016

Let LSI be the latus rectum, C be the centre of the ellipse and the coordinates of L be (x, y) then x = CS = 3 ae ...... (1)
And y = SL = b2/3a = (9a2(1 - e2))/3a = 3a (1 - e2) ...... (2)
Eliminating the variable 'e' from (1) and (2) we get the locus of L.
Hence putting the value of e from (1) and (2), we get:-
y = 3a(1-x2/9a2)
x2 = 9a2 - 3ay
x2 = 3a(3a -...  more»
Let LSI be the latus rectum, C be the centre of the ellipse and the coordinates of L be (x, y) then x = CS = 3 ae ...... (1)
And y = SL = b2/3a = (9a2(1 - e2))/3a = 3a (1 - e2) ...... (2)
Eliminating the variable 'e' from (1) and (2) we get the locus of L.
Hence putting the value of e from (1) and (2), we get:-
y = 3a(1-x2/9a2)
x2 = 9a2 - 3ay
x2 = 3a(3a - y), which is clearly a parabola. Similarly, we can show that the locus of L' is x2 = 3ay(y + 3a) which is again a parabola. «less

Tejas replied | 02/12/2016

Co-ordinate of extreme point of latus rectum would b (c,y).You can find ‘y’ by taking ‘x’ as ‘c’ in general equation,which comes out to b2/a.So the co-ordinate becomes {(a2-b2)1/2,b2/a}.Eliminating ‘b’ you will get a2-x2=ay.

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R

Rajini Sucharita 14/11/2016 in  Mathematics(Class XI-XII Tuition (PUC)),

What is the equation of chord joining two points on the hyperbola?

Sudeesh replied | 26/11/2016

xy1 + yx1 = 2c * c.

Ankit replied | 09/12/2016

xy1 + yx1 = 2c * c.

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R

Raghbendra 14/11/2016 in  Mathematics(Class XI-XII Tuition (PUC)),

What are the equation for finding the coordinates of the point of contact?

Rachna replied | 21/11/2016

To find the coordinates of point of contact you have to solve the equations simultaneously either by elimination method, or substitution or cross amulet up location method

Ankit replied | 09/12/2016

To find the coordinates of point of contact you have to solve the equations simultaneously either by elimination method, or substitution or cross amulet up location method.

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S

Sangeetha 14/11/2016 in  Mathematics(Class XI-XII Tuition (PUC)),

What is the difference between the diameter and conjugate diameter?

Nirmal replied | 03/12/2016

Hello Sangeetha,
Two diameters are conjugate if each bisects the chords parallel to the other diameter. If you consider the Hyperbola and Parabola, things get a little complicated. It is much simpler in Circles and Ellipses. Consider reading this link if you want to know more.

http:/...  more»
Hello Sangeetha,
Two diameters are conjugate if each bisects the chords parallel to the other diameter. If you consider the Hyperbola and Parabola, things get a little complicated. It is much simpler in Circles and Ellipses. Consider reading this link if you want to know more.

Regards

Nirmal «less

Rahul replied | 05/12/2016

Hello Sangeetha,
If we talk about ellipse, any chord that passes through the center of an ellipse is call its diameter. It follows that the family of parallel chords define two diameters: one in the direction to which they are all parallel and the other the locus of their midpoints. Such two diameters are called conjugate.

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S

Sita 14/11/2016 in  Mathematics(Class XI-XII Tuition (PUC)),

Prove that the midpoints of chords of the hyperbola (square of x)/(square of a)–(square of y)/(square of b) = 1 parallel to the diameter y = mx lie on the diameter (square of a)my = (square of b) x?

Ankit replied | 09/12/2016

Answer to this question cant be texted. Have a live session with me.

A

Aravind replied | 12/12/2016

Let the cord parallel to y=mx be
y=mx+c..............(1)
Then the points of intersection with the hyperbola are given by:-

x^2/a^2 - (mx+c)^2/b^2 = 1
or (1/a^2-m^2/b^2)x^2-2mxc/b^2-c^2/b^2-1=0 .....(2)

Let (x1,y1) and (x2,y2) be the end points of the cord.
Then x1 and x2 both satisfy Eq.(2)

So x1+x2 = [2mc/b^2]/(1/a^2-m^2/b^2)
...  more»
Let the cord parallel to y=mx be
y=mx+c..............(1)
Then the points of intersection with the hyperbola are given by:-

x^2/a^2 - (mx+c)^2/b^2 = 1
or (1/a^2-m^2/b^2)x^2-2mxc/b^2-c^2/b^2-1=0 .....(2)

Let (x1,y1) and (x2,y2) be the end points of the cord.
Then x1 and x2 both satisfy Eq.(2)

So x1+x2 = [2mc/b^2]/(1/a^2-m^2/b^2)
= 2mc/(b^2/a^2-m^2)...............(3)

Let (h,k) be the midpoint of the cord. Then
h=(x1+x2)/2 = mc/(b^2/a^2-m^2)
and k=(y1+y2)/2.

Now both (x1,y1) and (x2,y2) satisfy Eq(1) so that
k=mh+c. Substituting the value of c from Eq(3)
k=mh+(b^2/a^2-m^2)h/m = b^2/(a^2m)
Or a^2 mk = b^2h
Therefore the locus of the midpoint (h,k) of the cord
is a^2 my = b^2x. «less

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Poonam 14/11/2016 in  Mathematics(Class XI-XII Tuition (PUC)),

Prove that the angle subtended by any chord of a rectangular hyperbola at the centre is the supplement of the angle between the tangents at the end of the chord?

Ankit replied | 09/12/2016

Solve with the help of parametric equations of hyperbola.

Mohit replied | 14/12/2016

Solve with the help of parametric equations of hyperbola.

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B

Bhupen 14/11/2016 in  Mathematics(Class XI-XII Tuition (PUC)),

On a level plain the crack of the rifle and the thud of the ball striking the target are heard at the same instant, prove that the locus of the hearer is a hyperbola?

Jaya Laxmi Kl replied | 24/11/2016