Parabola Ques 56
- The tangent $P T$ and the normal $P N$ to the parabola $y^{2}=4 a x$ at a point $P$ on it meet its axis at points $T$ and $N$, respectively. The locus of the centroid of the triangle $P T N$ is a parabola, whose
(2009)
(a) vertex is $\frac{a}{3}, 0$
(b) directrix is $x=0$
(c) latusrectum is $\frac{2 p}{3}$
(d) focus is $(a, 0)$
Integer Answer Type Question
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Answer:
Correct Answer: 56.$(a, d)$
Solution:
Formula:
- 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
$$ [\beta y-2 a(x+\alpha)]^{2}=\left(\beta^{2}-4 a \alpha\right)\left(y^{2}-4 a x\right) $$
$\left[\because T^{2}=S \cdot S_{1}\right]$
$\Rightarrow \beta^{2} y^{2}+4 a^{2}\left(x^{2}+\alpha^{2}+2 x \cdot \alpha\right)-4 a \beta y(x+\alpha)$
$=\beta^{2} y^{2}-4 \beta^{2} a x-4 a \alpha y^{2}+16 a^{2} \alpha x$
$\Rightarrow \beta^{2} y^{2}+4 a^{2} x^{2}+4 a^{2} \alpha^{2}+8 x \alpha a^{2}$
$=\beta^{2} y^{2}-4 \beta^{2} a x-4 a \alpha y^{2}+16 a^{2} \alpha x-4 a \beta x y-4 a \beta \alpha y \ldots$ (i)
Now, coefficient of $x^{2}$ is $4 a^{2}$
coefficient of $x y=-4 a \beta$
coefficient of $y^{2}=4 a x$
Again, angle between the two of Eq. (i) is given as $45^{\circ}$
$$ \begin{array}{lc} \therefore & \tan 45^{\circ}=\frac{2 \sqrt{h^{2}-a^{2}b^{2}}}{a+b} \\ \Rightarrow & 1=\frac{2 \sqrt{h^{2}-a b}}{a+b} \\ $\Rightarrow & a+b=2 \sqrt{h^{2}-ab} \$ \Rightarrow & (a+b)^{2}=4\left(h^{2}-a^{2}\right) \ \Rightarrow & \left(4 a^{2}+4 a \alpha\right)^{2}=4\left[4 a^{2} \beta^{2}-\left(4 a^{2}\right)(4 a \alpha)\right] \\ \Rightarrow & 16 a^{2}(a+\alpha)^{2}=4 \cdot 4 a^{2}\left[\beta^{2}-4 a \alpha\right] \\ \Rightarrow & \alpha^{2}+6 a \alpha+a^{2}-\beta^{2}=0 \\ \Rightarrow & (\alpha+3 a)^{2}-\beta^{2}=8 a^{2} \end{array} $$
Thus, the required equation of the locus is $(x+3 a)^{2}-y^{2}=8 a^{2}$ which is a hyperbola.
BETA
