JEE Main Pyq of tangent to parabola - easy way out

[00:00:05]J E main PYQ of tangents to parabola.

[00:00:14]Question one. The equation of the common tangent to the curve y^2 = 8x and xy = -1.

[00:00:22]Tangent to the curve y^2 = to 8x is y = mx + 2 upon m. So it must satisfy xy = to minus1 as it is common tangent of both curves.

[00:00:36]x into mx + 2 upon m equals to minus1 which implies mx 2 + 2x upon m + 1 is equal to zero since it has equal roots.

[00:00:47]Therefore d=0.

[00:00:49]Thus 4 upon m² - 4 m will be equal to zero. This implies m cub= 1. m is equal to 1. Equation of common tangent is y = x + 2.

[00:01:05]The equations of two sides of a variable triangle are x=0 and y = 3 and its third side is a tangent to the parabola y^2= 6x. What is locus of circumcenter?

[00:01:20]Two sides of a triangle are x= to 0 and y = 3 comma third side is tangent to y^2 = 6x. The parametric coordinates of y^2 = to 6x are 6t ^2a 6t which is also the point of tangency.

[00:01:39]Tangent at any point x1 comma y1 on y^2 = 4 ax is y1 = 4 ax + x1 upon 2.

[00:01:49]Equation of tangent at point P 6 t ^2 6t on parabola is y into 6t = 6 into x + 6t ^2 upon 2.

[00:02:02]On further calculating we get 2 yt = x + 6 t ^ 2. Equation of tangent is x - 2 yt + 6t ^2 = 0.

[00:02:17]Intersection of tangent and side y = 3 gives vertex b. So put y = to 3 in the equation of tangent x - 2 yt + 6t ^2 = 0. On putting y = 3, we get x = 6 t - 6 t ^ 2. Thus vertex b is 6 t - 6 t ^2.

[00:02:43]Vertex C is intersection of side X equals to 0 and tangent. So put X=0 in equation of tangent X - 2 YT + 6 T ^²= 0. On putting X= 0 in the equation of tangent we get Y = 3T. Vertex C is 0a 3T.

[00:03:10]Triangle ABC is right angle triangle right angled at vertex A as sides X equals to 0 and Y equals to 3 are perpendicular to each other. Therefore circumcenter is midpoint of side BC.

[00:03:25]Let circumcenter be H comma K. As circumcenter is midpoint of side BC. H is equal to 6T - 6 T ^² upon 2 which is equal to 3T - 3T ^ 2. Let this be equation 1. And k is equal to 3 + 3t upon 2. Let this be equation 2.

[00:03:53]On solving equation 2, we get 2k - 3 = 3t which further gives t= 2kus 3 upon 3.

[00:04:05]On putting the value of t in equation 1 we get h = 3 * 2k - 3 upon 3 - 3 into 2k - 3 upon 3² h is equal to 6k - 9 upon 3 - 4k 2 - 12k + 9 upon 3. On further solving we get 3 h equal to 6k - 9 - 4k 2 - 12k + 9 which gives 4k 2 - 18 k + 3 h + 18 = 0.

[00:04:48]On replacing h comma k with x comm y we get 4 y ^2 - 18 y + 3x + 18 = 0 as the locus of circumcenter. Enter.