EllipseeasyFree

Ellipse: Conics Same Pair Foci Point Common Both Find

JEE Maths question with a full step-by-step solution.

Question

The conics

\dfrac{x^{2}}{a^{2}} + \dfrac{y^{2}}{b^{2}} = 1

and

\dfrac{x^{2}}{A^{2}} - \dfrac{y^{2}}{B^{2}} = 1

, where

a > A

and

a > b

, have the same pair of foci

S_1

and

S_2

, and

P

is a point common to both. Find the quadratic equation whose roots are

S_1P

and

S_2P

x^{2} + 2ax + \left(a^{2}-A^{2}\right) = 0

x^{2} - 2Ax + \left(a^{2}+A^{2}\right) = 0

x^{2} - 2Ax + \left(a^{2}-A^{2}\right) = 0

x^{2} - 2ax + \left(a^{2}-A^{2}\right) = 0

correct

Solution

Step 1:

P

lies on the ellipse, so

S_1P + S_2P = 2a

and

P

lies on the hyperbola with the same two foci, so

\left|S_1P - S_2P\right| = 2A

Step 2: Taking

S_1P - S_2P = 2A

S_1P = a + A, \qquad S_2P = a - A

Both are positive, as

a > A

. The other sign only interchanges the two lengths, so the quadratic is the same either way. Step 3:

\text{sum} = (a+A)+(a-A) = 2a, \qquad \text{product} = (a+A)(a-A) = a^{2}-A^{2}

Step 4:

x^{2} - (\text{sum})x + (\text{product}) = 0 \ \Longrightarrow\ x^{2} - 2ax + \left(a^{2}-A^{2}\right) = 0

Answer: (4).

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