Question

Given the parabola $y^2=4ax,$ find the pair of tangents from the point (2, 6) where a =2.

• A. $36x^2-4y^2-48xy+64x+64y=0$
• B. $36x^2-4y^2-8xy+64x+64y+64=0$
• C. $36x^2-4y^2-48xy+64x+64y+64=0$
• D. $36x^2-64y^2-48xy+64x+64y+64=0$

Answer 1

The correct option is C $36x^2-4y^2-48xy+64x+64y+64=0$

It is useful to rememberthe formula for pair of tangents from a point to a parabola which is $T^2=s{s}^{\prime }$

Where, $T=y{y}^{\prime }-2a(x+x^{\prime })$

$S=y^2-4ax$

$S^{\prime }=y^{\prime }2-4a{x}^{\prime }$

Here, $x^{\prime }=2,y^{\prime }=6,a=2$

$\therefore T^2=s{s}^{\prime }\Rightarrow \left(y\right(6)-2(2\left)\right(x+2){)}^2=y^2-4ax(6^2-4\left(2\right)\left(2\right))$

$\Rightarrow 36x^2-4y^2-48xy+64x+64y+64=0$