(i) If the normal to the parabola y^2=4 a x at the point (a t^2, 2 a t) meets the parabola again

(i) If the normal to the parabola y^2=4 a x at the point (a...

Key Concepts

Chord of a Conic Section

A chord is defined as a line segment with both endpoints on a conic section. In the context of parabolas, the chord formed by the intersection of a line (such as a normal) with the conic is analyzed for its length and geometric properties. Investigating chords involves combining algebraic methods with geometric insights, as well as optimizing their length or other attributes under certain conditions.

Intersection of a Line with a Parabola

This key concept deals with finding the points at which a given line, such as a normal, meets the parabola. The process typically involves substituting the line equation into the parabolic equation to yield a quadratic equation, whose roots correspond to the x or parameter values of the intersection points. Analyzing these intersections provides insights into the structure and symmetry of the parabola.

Optimization in Coordinate Geometry

Optimization involves finding the minimum or maximum values of geometric quantities, such as the length of a chord. In coordinate geometry, this is usually addressed by expressing the quantity in terms of a parameter and then using calculus, particularly differentiation, to determine the conditions under which the quantity is minimized or maximized. This method is essential for solving problems that ask for the shortest or longest segment under given constraints.

Normals to a Parabola

Normals are lines perpendicular to the tangent at a given point on the curve. For a parabola, the normal at any point can be derived from the derivative of the curve, and they often have interesting properties such as intersecting the parabola at other distinct points. Understanding the normal is crucial for exploring the geometric relations and optimizations associated with it.

Parabolic Parametric Representation

This concept involves expressing the points on a parabola using a parameter, which simplifies many geometric calculations. By representing the coordinates of a point on the parabola in terms of a single parameter, it becomes easier to derive relationships between points, tangents, and normals, and to solve for intersections between lines and the parabola.

Transcript

00:01 In this problem of conic section we have given that if the normal to the parabola whose equation is y is equal to 4 a x at the point say a t square and 2 a t meets the parabola again at a t1 square and twice of a t 1 then we have to prove that t 1 is equals to minus t is equals to this is minus t minus 2 so this would be minus t minus 2 divided with t so now first we would write the equation of the normal of parabola at at t square and 280 so this is given why y is equal to minus t x plus twice of a t plus a t plus a t cube plus a t cube now this is the normal and this passes through the point a t square plus two twice of 80 so we have to put x is equals to 81 square and y is equals to 88 so putting the value twice of a t1 and this is minus t multiplied with a t square plus 2 a t plus this would be a t cube and now from here a is cancelled out from all the terms so a is cancelled out and we can write it as twice of t1 and send it here so this would be plus t multiplied with t1 whole square is equals to now we are taking common t so here this would be t and here this is 2 plus t square and similarly we are taking common here t1 so this is t1 and here we left with 2 plus t1 is square and here this is t and 2 plus t2 square and further solving we are getting t1 is equals to minus t minus 2 divided with t so we can say hence proved hence proved and now second part so in second part we have to find the minimum length of such cord so here that means we have to find the minimum length of such cord so here that means we have to find the minimum length so length of code joining a t2 square and this is 80 square and twice of 80 and here second would be a t1 square and twice of 80 so here by distance formula we are getting the distance between them is say l is equals to here by distance formula we are getting 4a and 1 plus t is square and 3 divided with 2 divided with t squared and now l is minimum.

03:00 So here l is minimum when dl divided with dt is equals to 0.

03:09 So we have to take the differentiation...

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