The tangents at the points A(1,3) and B(1,-1) on the parabola y^2-2 x-2 y=1 meet at the po…

Question

The tangents at the points $A(1,3)$ and $B(1,-1)$ on the parabola $y^{2}-2 x-2 y=1$ meet at the point $P$. Then the area (in unit ${ }^{2}$ ) of the triangle $P A B$ is :

Accepted Answer

Given curve : ${y^2} - 2x - 2y = 1$.

Can be written as

${(y - 1)^2} = 2(x + 1)$

And, the given information can be plotted as shown in figure

Tangent at $A:2y - x - 5 = 0$ {using $T = 0$}

Intersection with $y = 1$ is $x =  - 3$

Hence, point P is $( - 3,1)$

Taking advantage of symmetry

Area of $$\Delta PAB = 2 \times {1 \over 2} \times (1 - ( - 3)) \times (3 - 1)$$

= 8 sq. units

Solutions · 1