Q94 / 160 · AP EAMCET 2024 (19 May Shift 2) · Mathematics

The locus of the mid points of the chords of the hyperbola $x^2-y^2=a^2$ which touch the parabola $y^2=4 a x$ is

  • Correct answer
    A
    $x\left(y^2-x^2\right)=a y^2$
  • B
    $x\left(x^2+y^2\right)=y^2+x$
  • C
    $a x^3+y^3=3 x$
  • D
    $x\left(x^2-y^2\right)=a^2$

Solution

Let mid point of the chord of the given parabola is $(\alpha, \beta)$ $\Rightarrow$ equation of chord is $x x_1-y y_1=x_1^2-y_1^2...(i)$ $\Rightarrow \alpha x-\beta y=\alpha^2-\beta^2 \Rightarrow y=\frac{\beta^2-\alpha^2}{\beta}+\frac{\alpha x}{\beta}$ Now, this equation touches to the parabola $y^2=4 a x$ So, $-\frac{\left(\alpha^2-\beta^2\right)}{\beta}=\frac{a \beta}{\alpha}$ $\Rightarrow-\left(x^2-y^2\right) x=a y^2 \Rightarrow\left(y^2-x^2\right) x=a y^2$

Hence, the correct answer is option: 'A'.

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