Find-the-cordinates-of-the-two-points-on-the-curve-y-4-x-2-whose-tangents-pass-through-the-point-1-7-

Question Number 5614 by Rasheed Soomro last updated on 22/May/16

Find the cordinates of the two points

on the curve y = 4 - x^{2} whose tangents

pass through the point (- 1, 7) .

Commented by Rasheed Soomro last updated on 24/May/16

T h^{α} {n k}^{S}!

Commented by Yozzii last updated on 23/May/16

L e t t h e t a n g e n t t o t h e c u r v e b e g i v e n

b y y - y_{0} = \frac{d y}{d x} ∣_{x = m} (x - x_{0}) w h e r e

(x_{0}, y_{0}) = (- 1, 7), \frac{d y}{d x} = \frac{d}{d x} (4 - x^{2}) = - 2 x

a n d x = m c o r r e s p o n d s t o a p o i n t

o f t a n g e n c y w i t h x = m .

∴ y - 7 = (- 2 m) (x + 1)

A t t h e p o i n t o f t a n g e n c y, y = 4 - m^{2} .

∴ 4 - m^{2} - 7 = - 2 m (m + 1)

- 3 - m^{2} = - 2 m^{2} - 2 m

m^{2} + 2 m - 3 = 0

(m + 3) (m - 1) = 0

m = 1, - 3 \Rightarrow y = 3, - 5 .

∴ t h e r e q u i r e d p o i n t s a r e (1, 3), (- 3, - 5) .

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