Show that the equation of the normal to the ellipse `(x^(2))/(25)+(y^(2))/(9)=1 "at" ((5)/(sqrt(2)),(3)/(sqrt(2)))` is the line `5x-3y=8sqrt(2)`.

Find the equation of the normal to the circle x^2+y^2=9 at the point (1/(sqrt(2)),1/(sqrt(2))) .

Then angle of intersection of the normal at the point (-5/(sqrt(2)),3/(sqrt(2))) of the curves x^2-y^2=8 and 9x^2+25 y^2=225 is 0 (b) pi/2 (c) pi/3 (d) pi/4

A beam of light is sent along the line x-y=1 , which after refracting from the x-axis enters the opposite side by turning through 30^0 towards the normal at the point of incidence on the x-axis. Then the equation of the refracted ray is (a) (2-sqrt(3))x-y=2+sqrt(3) (b) (2+sqrt(3))x-y=2+sqrt(3) (c) (2-sqrt(3))x+y=(2+sqrt(3)) (d) y=(2-sqrt(3))(x-1)

Find the equation of normal to the ellipse x^(2)+4y^(2)=4 at (2 cos theta, sin theta) . Hence find the equation of normal to this ellipse which is paralel to the line 8x+3y=0 .

The number of solutions of the equation sqrt(x^(2))-sqrt((x-1)^(2))+sqrt((x-2)^(2))=sqrt(5) is

The slopes of the common tangents of the ellipse (x^2)/4+(y^2)/1=1 and the circle x^2+y^2=3 are +-1 (b) +-sqrt(2) (c) +-sqrt(3) (d) none of these

Let the equation of an ellipse be (x^(2))/(144) + (y^(2))/(25) = 1 . Then the radius of the circle with centre (0 , sqrt(2)) and passing through the foci of the ellipse is _

Let the equation of an ellipse be (x^(2))/(144)+(y^(2))/(25)=1 . Then the radius of the circle with centre (0, sqrt(2)) and passing through the foci of the ellipse is

If sqrt(1-x^(2))+sqrt(1-y^(2))=a(x-y) , show that, (dy)/(dx)=(sqrt(1-y^(2)))/(sqrt(1-x^(2))) .