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Question: Two tangents are drawn from the point (–2, –1) to the parabola \(\tan \alpha\)=...

Two tangents are drawn from the point (–2, –1) to the parabola tan⁡α\tan \alpha=

A

3

B

1/3

C

2

D

½

Answer

3

Explanation

Solution

Equation of pair of tangent from (−2,−1)( - 2 , - 1 )to the parabola is given by SS1=T2S S _ { 1 } = T ^ { 2 } i.e. (y2−4x)(1+8)=[y(−1)−2(x−2)]2\left( y ^ { 2 } - 4 x \right) ( 1 + 8 ) = [ y ( - 1 ) - 2 ( x - 2 ) ] ^ { 2 }

⇒ 9y2−36x=[−y−2x+4]29 y ^ { 2 } - 36 x = [ - y - 2 x + 4 ] ^ { 2 }

⇒9y2−36x=y2+4x2+16+4xy−16x−8y9 y ^ { 2 } - 36 x = y ^ { 2 } + 4 x ^ { 2 } + 16 + 4 x y - 16 x - 8 y

⇒ 4x2−8y2+4xy+20x−8y+16=04 x ^ { 2 } - 8 y ^ { 2 } + 4 x y + 20 x - 8 y + 16 = 0

∴ tan⁡α=∣2h2−aba+b∣\tan \alpha = \left| \frac { 2 \sqrt { h ^ { 2 } - a b } } { a + b } \right| =∣24−4(−8)4−8∣=∣12−4∣=3= \left| \frac { 2 \sqrt { 4 - 4 ( - 8 ) } } { 4 - 8 } \right| = \left| \frac { 12 } { - 4 } \right| = 3