Tangent at any point on the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ cut the axis at A and B respectively. If the rectangle OAPB (where O is origin) is completed then locus of point P is given by

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A tangent to the ellipse x2a2+y2b2=1,(a>b) having slope 1 intersects the axis of x & y in point A & B respectively. If O is the origin then find the area of △OAB.

The correct option is D 12(a2+b2)equation of tangent is y=x±√a2+b2X-intercept=∓√a2+b2 and Y-intercept=±√a2+b2Therefore, area of △OAB=12(a2+b2)

A tangent to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, (a > b)$ having slope $1$ intersects the axis of $x$ & $y$ in point $A$ & $B$ respectively. If $O$ is the origin then find the area of $△ O A B$ .

The correct option is D $\frac{1}{2} (a^{2} + b^{2})$
equation of tangent is $y = x \pm \sqrt{a^{2} + b^{2}}$
X-intercept $= \mp \sqrt{a^{2} + b^{2}}$ and Y-intercept $= \pm \sqrt{a^{2} + b^{2}}$
Therefore, area of $△ O A B = \frac{1}{2} (a^{2} + b^{2})$