JEE PYQ: Hyperbola Question 1
Question 1 - 2021 (16 Mar Shift 1)
The locus of the midpoints of the chord of the circle, $x^2 + y^2 = 25$ which is tangent to the hyperbola, $\frac{x^2}{9} - \frac{y^2}{16} = 1$ is:
(1) $(x^2 + y^2)^2 - 16x^2 + 9y^2 = 0$
(2) $(x^2 + y^2)^2 - 9x^2 + 144y^2 = 0$
(3) $(x^2 + y^2)^2 - 9x^2 - 16y^2 = 0$
(4) $(x^2 + y^2)^2 - 9x^2 + 16y^2 = 0$
Show Answer
Answer: (4)
Solution
Let midpoint of chord be $(h, k)$. Equation of chord: $hx + ky = h^2 + k^2$, i.e. $y = -\frac{h}{k}x + \frac{h^2+k^2}{k}$. For this to be tangent to the hyperbola: $c^2 = a^2m^2 - b^2$, so $\left(\frac{h^2+k^2}{k}\right)^2 = 9\left(-\frac{h}{k}\right)^2 - 16$. This gives $(x^2+y^2)^2 = 9x^2 - 16y^2$, i.e. $(x^2+y^2)^2 - 9x^2 + 16y^2 = 0$.
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