Question

If the line $y=mx+c$ is a tangent to the parabola $y^2=4a(x+a)$, then

• A. $c=m^{2}a+\frac{a}{m}$
• B. $c=ma+\frac{a}{m}$
• C. $c=\frac{(m^{2}a-a^2)}{m}$
• D. $c=ma-\frac{a}{m}$

Answer 1

The correct option is B $c=ma+\frac{a}{m}$

Equation of tangent to the parabola $y^2=4a(x+a)$ at a point $\left(x_1{y}_1\right)$ on the curve is given by

$y{y}_1=4a(\frac{x+x_1}{2}+h)$

in our case the given equation of parabola is $y^2=4a(x+a)$

$\therefore$ the equation of tangent at a point $\left(x_1{y}_1\right)$ on the parabola is $y{y}_1=4a(\frac{x+x_1}{2}+0)$

$\Rightarrow 2y{y}_1=4ax+4a{x}_1+8a^2$

$\Rightarrow y=\frac{2a}{y_1}x+\frac{2a{x}_1+4a^2}{y_1}......\left(1\right)$

Given that the equation of tangent is

$\Rightarrow y=mx+c.........\left(2\right)$

comparing $\left(1\right)$ and $\left(2\right)$ we get

$\Rightarrow m=\frac{2a}{y_1}$ and $c=\frac{2a{x}_1+4a^2}{y_1}$

Now, $c=\frac{2a}{y_1}x+\frac{2a}{y_1}2a$

$\Rightarrow c=mx+2ma.........\left(3\right)$

Also the point $\left(x_1{y}_1\right)$ lies on the parabola $y^2=4a(x+a)$

$\therefore y_1^2=4a(x_1+a)$

$\Rightarrow \frac{y_1^2}{4a}-a=x_1$

$\Rightarrow \frac{y_1^2}{{\left(2a\right)}^2}-a=x_1$

$\Rightarrow \frac{a}{m^2}-a=x_1......\left(4\right)$

putting $\left(4\right)$ in $\left(3\right)$ we get,

$\Rightarrow c=m(\frac{a}{m^2}-a)+2am$

$\Rightarrow c=\frac{a}{m}+am$

Hence, solved.