# Find the equations of tangents to the curve $3x^2-y^2=8$,which pass through the point$\big(\frac{4}{3},0\big)$. (Note: This question has been split into 2 questions)

(Note: This question has been split into 2 questions) This question appeared in 65-1,65-2 and 65-3 versions of the paper in 2013.

## 1 Answer

Given $y=x^2\; and\; y=x^2$
$x^2=y$ represents a parabola with vertex (0,0),open upwards, in the positive direction.
$y=x,$ represents a line passing through the origin making an angle $45^{\circ}$ in the positive direction of x-axis and
$y=-x,$ represents a line passing through the origin making an angle $135^{\circ}$ in the positive direction of x-axis
Required area=2(shaded area in the I quadrent)
$A=2\int \limits_0^1 (x-x^2)dx$
$=2 \int \limits_0^1 x-2 \int \limits_0^1 x^2 dx$
$=2 \bigg[\frac{x^2}{2}-\frac{x^3}{3} \bigg]_0^1$
$=2\bigg[\frac{1}{2}-\frac{1}{3}\bigg]=2 \times \frac{1}{6}=\frac{1}{3} sq.units$
answered Mar 21, 2013 by
edited Mar 26, 2013

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