Let the line $$y=mx$$ and the ellipse $$2x2+y2=1$$ intersect at a point P in the first quadrant. the area of the traingle formed by the normal to this ellipse at P with axes is Normal is passing through the point $$-132$$,$$0$$

Question

Let the line $$y=mx$$ and the ellipse $$2x2+y2=1$$ intersect at a point P in the first quadrant.  the area of the traingle formed by the normal to this ellipse at P with axes is  Normal is passing through the point $$-132$$,$$0$$

Infinity Learn · Accepted Answer

Let P be $$x1$$,$$y1$$ Equation of normal at P is $$x2x1-yy1=-12$$ It passes through $$-132$$,$$0$$⇒$$-162x1=-12$$⇒$$x1=132$$ So $$y1=223$$ $$(as$$ P lies in $$1st$$  quadrant  The area of the triangle formed by this normal with axes is $$1223132Therefore$$ the area of the triangle is $$118square$$ units

$Let the line y = m x and the ellipse 2 x^{2} + y^{2} = 1 intersect at a point P in the first quadrant. the area of the traingle formed by the normal to this ellipse at P with axes is (N o r m a l i s p a s \sin g t h r o u g h t h e p o i n t (- \frac{1}{3 \sqrt{2}}, 0))$

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Let the line y=mx and the ellipse 2x2+y2=1 intersect at a point P in the first quadrant. the area of the traingle formed by the normal to this ellipse at P with axes is Normal is passing through the point -132,0

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$Let the line y = m x and the ellipse 2 x^{2} + y^{2} = 1 intersect at a point P in the first quadrant. the area of the traingle formed by the normal to this ellipse at P with axes is (N o r m a l i s p a s \sin g t h r o u g h t h e p o i n t (- \frac{1}{3 \sqrt{2}}, 0))$