Parabola Ques 10
- The angle between a pair of tangents drawn from a point $P$ to the parabola $y^2=4 a x$ is $45^{\circ}$. Show that the locus of the point $P$ is a hyperbola.
(1998, 8M)
Show Answer
Solution: Let $P(\alpha, \beta)$ be any point on the locus. Equation of pair of tangents from $P(\alpha, \beta)$ to the parabola $y^2=4 a x$ is
${[\beta y-2 a(x+\alpha)]^2=\left(\beta^2-4 a \alpha\right)\left(y^2-4 a x\right)} $ $\quad \left[\because T^2=S \cdot S_1\right] $
$=\quad \beta^2 y^2+4 a^2\left(x^2+\alpha^2+2 x \cdot \alpha\right)-4 a \beta y(x+\alpha) $
$\Rightarrow \quad \beta^2 a x-4 a \alpha y^2+16 a^2 \alpha \alpha x $
$= \quad \beta^2 y^2-4 a^2 x^2+4 a^2 \alpha^2+8 x \alpha a^2$
$=\quad \beta^2 y^2-4 \beta^2 a x-4 a \alpha y^2+16 a^2 \alpha x-4 a \beta x y-4 a \beta \alpha y \ldots \text { (i) } $
Now, coefficient of $x^2=4 a^2$
coefficient of $x y=-4 a \beta$
coefficient of $y^2=4 a \alpha$
Again, angle between the two of Eq. (i) is given as $45^{\circ}$
$\therefore \quad \tan 45^{\circ}=\frac{2 \sqrt{h^2-a b}}{a+b}$
$\Rightarrow \quad 1=\frac{2 \sqrt{h^2-a b}}{a+b}$
$\Rightarrow \quad a+b=2 \sqrt{h^2-a b}$
$\Rightarrow \quad(a+b)^2=4\left(h^2-a b\right)$
$\Rightarrow \quad \left(4 a^2+4 a \alpha\right)^2=4\left[4 a^2 \beta^2-\left(4 a^2\right)(4 a \alpha)\right]$
$\Rightarrow \quad 16 a^2(a+\alpha)^2=4 \cdot 4 a^2\left[\beta^2-4 a \alpha\right]$
$\Rightarrow \quad \alpha^2+6 a \alpha+a^2-\beta^2=0$
$\Rightarrow \quad(\alpha+3 a)^2-\beta^2=8 a^2$
Thus, the required equation of the locus is $(x+3 a)^2-y^2=8 a^2$ which is a hyperbola.
BETA
