Show that for any point of the curve `x^2 - y^2 = a^2` the segment of the normal from the poin
34 views · Published 13 October 2018 · 5:58 · Indexed 25 September 2026
Channel: Doubtnut · 2018 · Education
To ask Unlimited Maths doubts download Doubtnut from - https://goo.gl/9WZjCW Show that for any point of the curve `x^2 - y^2 = a^2` the segment of the normal from the point to the point of intersection of the normal with the x-axis is equal to the distance of the point from the origin.
More from this channel
-
4:29
IIT JEE LIMITS AND DERIVATIVES Evaluate the limit:
`("lim")_(xvecoo)((x+2)/(x+1))^(x+3)`
-
3:50
`A D` and `B E` are respectively altitudes of an isosceles triangle `A B C` with `A C=B C` . Pro...
-
4:19
IIT JEE PERMUTATIONS AND COMBINATIONS Find the number of distinct rational numbers `x`
such that ...
-
2:15
IIT JEE PERMUTATIONS AND COMBINATIONS Find the sum of the series `sum_(r=1)^n rxxr !dot`
-
2:47
JEE MAINS 2018 Find the points of local maxima and local minima,
if any, of each of the follo...
-
3:03
Find the number of polynomials of the form `x^3+a x^2+b x+c`
that are divisible by `x^2+1,w h e...
-
3:16
IIT JEE PERMUTATIONS AND COMBINATIONS Find the number of ways in which 6 boys and 6 girls can be ...
-
2:26
IIT JEE COMPLEX NUMBERS AND QUADRATIC EQUATIONS Prove that graphs of `y=x^2+2a n dy=3x-4`
never i...