Question

The locus of poles of chords of the parabola $y^2=4ax$ which subtend a constant angle $\alpha$ at the vertex of the parabola is:

• A. $(x+4a{)}^2{\tan }^2\alpha =4(y^2-4ax)$
• B. $(x+a{)}^2{\tan }^2\alpha =y^2-4ax$
• C. $(x+a{)}^2{\cot }^2\alpha =y^2-4ax$
• D. $(x+4a{)}^2{\cot }^2\alpha =y^2-4ax$

Answer 1

Answer: (A)

$(x+4a{)}^2{\tan }^2\alpha =4(y^2-4ax)$

Let $P(x_1,y_1)$ be the midpoint of the chord $BC$ of the parabola $y^2=4ax$
Equation of $BC$ is $y{y}_1-2a(x+x_1)=y_1^2-4a{x}_1$
$y{y}_1-2ax=y_1^2-2a{x}_1$
Join AB, AC given that $\angle BAC=\alpha$
Homogenising $y^2=4ax$
$(y^2-4ax)\frac{(y{y}_1-2ax)}{(y_1^2-2ax)}=0$
$y^2(y_1^2-2a{x}_1)-4a{y}_{1}xy+8a^2{x}^2=0$
$\tan \alpha =\frac{2\sqrt{4a^2{y}_1^2-8a^2(y_1^2-2a{x}_1)}}{y_1^2-2a{x}_1+8a^2}$
Substituting $y_1^2=4a{x}_1$ in the denominator and cross multiplying and squaring both side, we get
${\tan }^2\alpha (2a{x}_1+8a^2{)}^2=|4(16a^3{x}_1-4a^2{y}_1^2)|$
${\tan }^2\alpha (2a{x}_1+8a^2{)}^2=16a^2(4a{x}_1-y_1^2)$
Simplifying other,
${(x_1+4\alpha )}^{2}ta{n}^2\alpha =4(y_1^2-4a{x}_1)$
Therefore, Locus of $P$ is
$(x+4a{)}^2{\tan }^2\alpha =4(y^2-4ax)$