Question

What's the equation of a tangent on the parabola
$y^2=5x$ at the point (5, 5)

• A. x + 2y + 5 = 0
• B. x - 5y + 2 = 0
• C. x - 2y + 5 = 0
• D. x + 5y - 2 = 0

Answer 1

The correct option is C

x - 2y + 5 = 0

Method 1

Given equation of parabola is,

$y^2=5x$ .............(1)

The tangent will be having a slope equal to ${\begin{array}{c}\frac{dy}{dx}\end{array}|}_{(5,5)}$

and also passes through the point (5,5)

$\therefore$ differentiating (1) with respect to x

$2y.\frac{dy}{dx}=5$

${\begin{array}{c}\frac{dy}{dx}=\frac{5}{2y}\end{array}|}_{(5,5)}$$=\frac{1}{2}$

$\therefore$ equation of the tangent

$y-5=\frac{1}{2}(x-5)$

2y-10=x-5

x-2y+5=0

Method 2

Here we can use the formula $T_1=0$ for tangent

at the point $(x_1,y_1).T_1$ is obtained by replacing

$-x^2\text{with}x{x}_1$

$-y^2\text{with}y{y}_1$

$-x\text{with}\frac{x+x_1}{2}$

$-y\text{with}\frac{y+y_1}{2}$

$\therefore T_1=0$
$\Rightarrow y{y}_1=5.\left[\frac{x+x_1}{2}\right]$

$\Rightarrow 5y=\frac{5.(x+5)}{2}$

$\Rightarrow 2y=x+5$

$\Rightarrow x-2y+5=0$