Question

If the normal to the curve y² = 5x -1, at the point(1, -2) is of the form ax - 5y + b = 0, then a + b = _________________.

Answer 1

The equation of given curve is

y² = 5x −1

Differentiating both sides with respect to x, we get

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

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

∴ Slope of normal at (1, −2)

$=-\frac{1}{{\left({\displaystyle \frac{dy}{dx}}\right)}_{\left(1,-2\right)}}$

$=-\frac{1}{{\displaystyle \frac{5}{2\times \left(-2\right)}}}$

$=\frac{4}{5}$

So, the equation of normal at (1, −2) is

$y-\left(-2\right)=-\frac{1}{{\left({\displaystyle \frac{dy}{dx}}\right)}_{\left(1,-2\right)}}\left(x-1\right)$

$\Rightarrow y-\left(-2\right)=\frac{4}{5}\left(x-1\right)$

$\Rightarrow 5y+10=4x-4$

$\Rightarrow 4x-5y-14=0$

Comparing with the given equation of normal ax − 5y + b = 0, we get

a = 4 and b = −14

∴ a + b = 4 + (−14) = −10

Thus, the value of a + b is −10.

If the normal to the curve y² = 5x −1, at the point(1, −2) is of the form ax − 5y + b = 0, then a + b = ___−10___.