If the normal at an end of a latus rectum of an ellipse...

If the normal at an end of a latus rectum of an ellipse passes through an extremity of the minor axis, show that the eccentricity e of the ellipse is a solution of the equation $\mathrm{e}^{4}+\mathrm{e}^{2}-1=0$.

Key Concepts

Ellipse

An ellipse is a conic section defined as the locus of points where the sum of the distances from two fixed points (the foci) is constant. Its geometric properties, such as semi-major and semi-minor axes, foci, and symmetry, play crucial roles in many derivations and applications in analytical geometry.

Eccentricity

The eccentricity of an ellipse is a dimensionless number that measures its deviation from being circular, defined as the ratio of the distance from the center to one of the foci over the semi-major axis. It is a key parameter that determines the shape of the ellipse, with values ranging between 0 (a circle) and 1 (a parabola in the limiting case).

Latus Rectum

The latus rectum is a line segment, drawn through a focus of an ellipse and perpendicular to the major axis, with both of its endpoints lying on the ellipse. It is intimately connected with the focal properties of the curve and often appears in discussions relating the conic’s dimensions and its eccentricity.

Normal to a Curve

The normal to a curve at a given point is the straight line perpendicular to the tangent at that point. In geometric problems involving conic sections, analyzing the properties and equations of the normal can yield important relationships, especially when geometric constraints (such as passing through specific points) are imposed.

Transcript

00:01 In this problem of conic section we have given that if the normal at an end of let us select term of an ellipse passes through an extremity of the minor axis.

00:12 We have to show that eccentricity e of the ellipse is a solution of e to the power 4 plus e square minus 1 is equal to 0.

00:24 So first we are considering let the ellipse be.

00:30 So this is the equation of ellipse which is x squared.

00:33 With a square plus y squared divided with b is equal to 1.

00:40 Now lb which is the end of latest rectum has coordinate so l this is the point n is the end of later st rectum then the coordinate of l is say this would be a e and a multiplied with 1 minus a square and now from here we know that this is a cos theta is equal to ae and from here we can say cost theta is b equals to a this is e actually and we know that sine theta so sine is square theta or we can say sine theta is equal to under root of 1 minus e square now normal at theta so here normal at point theta on the ellipse is a x divided with e minus b y divided with under root of 1 minus is square is equal to a square minus b square now here we can solve and we know that b is equal to a square multiplied with 1 minus e square so from here a is equal to this would be b divided with under root of 1 minus e square so putting this value so this would be a x divided with e minus a y so this is a y is equal to a square minus b square and from here we can cancel one term and b would be putting this term so we can put this value so b square is equal to this would be a square and plus this would be plus a square divided with e squared so this is cancelled out and a is also cancelled out x divided with e minus y is equals to a e square...

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