If the normal to the ellipse "`(x^(2))/(18)+(y^(2))/(8)=1`" at point (3,2)" is `"ax-by-c=0"` .(where "gcd(a,b,c)=1)" .Then the value of `a+b+c=?`

Find the normal to the ellipse (x^(2))/(18)+(y^(2))/(8)=1 at point (3,2).

If the normal to the ellipse (x)/(18)+(y)/(8)=1 at point (3,2) is ax-by-c=0 .(where gcd(a,b,c)=1) .Then the value of a+b+c ?

The line y=mx+c is a normal to the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1, if c

If y = x + c is a normal to the ellipse x^2/9+y^2/4=1 , then c^2 is equal to

If y=mx+c is a tangent to the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 then the point of contact is

A tangent is drawn to the ellipse to cut the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 and to cut the ellipse (x^(2))/(c^(2))+(y^(2))/(d^(2))=1 at the points P and Q. If the tangents are at right angles,then the value of ((a^(2))/(c^(2)))+((b^(2))/(d^(2))) is

If x^(2)y+y^(3)=2 then the value of (d^(2)y)/(dx^(2)) at the point (1,1) is (-(a)/(b)) where g.c.d(a, b)=1 then a+b=

If the line (ax)/(3)+(by)/(4)= cbe a normal to the ellipse (x^(2))/(a^(2))+(y^(2))/(a^(2))=1, show that 5c=a^(2)e^(2) where e is the eccentricity of the ellipse.

If normal at any point P to the ellipse +(x^(2))/(a^(2))+(y^(2))/(b^(2))=1,a>b meets the axes at M and N so that (PM)/(PN)=(2)/(3), then value of eccentricity e is