The normal to the curve y (x – 2) (x – 3) = x + 6 at the point where the curve intersects the y‑axis passes through the point

Engineering

Mathematics

Tangent and Normal

Question

$(\frac{1}{2}, \frac{1}{2})$

$(\frac{- 1}{2}, \frac{- 1}{2})$

$(\frac{1}{2}, \frac{- 1}{3})$

$(\frac{1}{2}, \frac{1}{3})$

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

Given curve y × (x – 2) (x – 3) = (x + 6) …….(1)

Let required point is P, where curve intersects y‑axis.

For P, put x = 0 ⇒ y = 1, so P(0, 1)

Equation (1) differentiate w.r.t. x

(x – 2) (x – 3) $\frac{dy}{dx}$ + y (2x – 5) = 1

$\Rightarrow {(\frac{dy}{dx})}_{(0, 1)} = 1$

Slope of normal = – 1

Equation of normal at P (0, 1) ⇒ (y – y₁) = m (x – x₁)

y – 1 = – 1 (x – 0) ⇒ x + y = 1 …….(2)

By option $(\frac{1}{2}, \frac{1}{2})$ will satisfy x + y = 1.