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+1 Mathematics

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Physics Teacher

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Educator with26+yrs of international exp.Teaching-Math,Science,Engg,Medical,EntranceExams

Dear Archit, you must have your basics right about the conics section, here particularly, Hyperbola. Suggestions: Try reading basics (Even though you know), then solve some ques based on it and then take on the particular questions. Devote time separately to such topics. I m sure your will over come...
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Dear Archit, you must have your basics right about the conics section, here particularly, Hyperbola. Suggestions: Try reading basics (Even though you know), then solve some ques based on it and then take on the particular questions. Devote time separately to such topics. I m sure your will over come by following the said pattern. Come back if you need more help. read less
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Hello Archit, Please see the solutions below: 2. Let (h,k) be the midpoint of the chord. Then the equation of the chord with midpoint (h,k) is: (hx/3)-(ky/7) = (h^2/3) - (k^2/7) (Equation given by T1 = S1). 7x + y = 20 is the equation of the same line. Comparing the coefficients of the line...
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Hello Archit, Please see the solutions below: 2. Let (h,k) be the midpoint of the chord. Then the equation of the chord with midpoint (h,k) is: (hx/3)-(ky/7) = (h^2/3) - (k^2/7) (Equation given by T1 = S1). 7x + y = 20 is the equation of the same line. Comparing the coefficients of the line we get: h/(3*7) = -k/(7) = ((h^2/3) - (k^2/7))/20 So k/h = -1/3 Now the line whose equation we require to find passes through the origin and (h,k) So the equation of the line is y = (k/h).x Since k/h = -1/3, the equation of the line is y = (-1/3)x ---------------------------------------------------------------------------------------------------------------------------------------------- 3. The equation of the hyperbola can be written as (x^2/16) - (y^2/25) = 1 Let the point on the Hyperbola be (4secA, 5 tan A) The equations of the assymptotes are given by (x^2/16) - (y^2/25) = 0 Factorizing out the 2 stratight lines, we get the 2 equations as y = 5x/4 and y = -5x/4 Let Q be the point on y = 5x/4. Since the abscissa remains same, the x coordinate of Q is 4secA. So the y coordinate of Q is y = 5.4secA/4 = 5secA So Q: (4secA, 5secA). Now R lies on y = -5x/4. Solving similarly like we did for Q, the coordinates of R would be (4secA, -5secA). We also have P: (4secA,5tanA) So, PQ = 5(secA-tanA) and PR = 5(secA+tanA) Therefore, PQ.PR = 25.((secA)^2-(tanA)^2) For any angle A, ((secA)^2-(tanA)^2) = 1 Therefore, PQ.PR = 25. Let me know if you have any questions. Thanks read less
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Its basic.....eqn. of chord of hyperbola - T1=S which is a standard representation of the chord of a hyperbola if x1,y1 r midpoints. from where u can find out slope of the chord in terms of x1and y1. now this chord is also tangent to the 2nd hyperbola .So put x1,y1 in the tangent eqn. and u also know...
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Its basic.....eqn. of chord of hyperbola - T1=S which is a standard representation of the chord of a hyperbola if x1,y1 r midpoints. from where u can find out slope of the chord in terms of x1and y1. now this chord is also tangent to the 2nd hyperbola .So put x1,y1 in the tangent eqn. and u also know the slope m which will b required. Thus u get a eqn. which contains x1,y1 as only variables . Thus u can find out its locus. But i suggest u to learn the topic logically before proceeding with this bcoz it was simple if u had complete idea of the thing read less
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(1+?2)/?2
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1/root 2 + 1
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Mathematics teacher

I can help you to solve these online
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JEE/Board Maths Tutor

i suggest u to learn the topic logically before proceeding with this bcoz it was simple if u had complete idea of the thing
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