The equation to the pair of tangents drawn from -1-2 to parabola x 2-2 y isA- 4 x2-2 x y-y2-4 x-6 y-4-0- x2-2 x y-3 y2-10 x-6 y-9-0- 4 x2-2 x y-3 y2-4 x-6 y-4-0D- x2-2 x y-3 y2-10 x-6 y-9-0

Byju's Answer

Standard XII

Mathematics

Pair of Tangents from an External Point

The equation ...

Question

The equation to the pair of tangents drawn from (–1, –2) to parabola $x^{2} = 2 y$ is

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A
Equation of pair of tangents is given by $T^{2} = S S_{1}$
Equation of pair of tangents from $(- 1, - 2) i s [- x - (y - 2)^{2}] = (x^{2} - 2 y) (1 + 4)$
$\Rightarrow (- x - y + 2)^{2} = 5 (x^{2} - 2 y) \Rightarrow 4 x^{2} - y^{2} - 2 x y + 4 x - 6 y - 4 = 0$

Suggest Corrections

Similar questions

The angle between a pair of tangents drawn from a point P to the circle $x^{2} + y^{2} + 4 x - 6 y + 9 s i n^{2} α + 13 c o s^{2} α = 0 i s 2 α .$ The equation of the locus of the point P is

Q. For all real value of x and y ,

x^{2} + 2 x y + 3 y^{2} - 6 y \geq \frac{- 9}{2}

The equation of the circle passing through the point (2,-1) and having two diameters along the pair of lines $2 x^{2} + 6 y^{2} - x + y - 7 x y - 1 = 0 i s$

Q. The equation of the parabola whose focus lies at the intersection point of the lines

x + y = 3

and

x - y = 1

and directrix is

x - y + 5 = 0

Q. The angle between the pair of tangents drawn from a point P to the circle

x^{2} + y^{2} + 4 x - 6 y + 9

s i n^{2} α + 13 c o s^{2} α = 0

2 α

. Then the equation of the locus of the point P is

The equation to the pair of tangents drawn from (–1, –2) to parabola x2=2y is

The correct option is A Equation of pair of tangents is given by T2=SS1 Equation of pair of tangents from (−1,−2)is[−x−(y−2)2]=(x2−2y)(1+4) ⇒(−x−y+2)2=5(x2−2y)⇒4x2−y2−2xy+4x−6y−4=0

The equation to the pair of tangents drawn from (–1, –2) to parabola $x^{2} = 2 y$ is

The correct option is A
Equation of pair of tangents is given by $T^{2} = S S_{1}$
Equation of pair of tangents from $(- 1, - 2) i s [- x - (y - 2)^{2}] = (x^{2} - 2 y) (1 + 4)$
$\Rightarrow (- x - y + 2)^{2} = 5 (x^{2} - 2 y) \Rightarrow 4 x^{2} - y^{2} - 2 x y + 4 x - 6 y - 4 = 0$