Normal of a Curve y = f(x)

Q. Let the normals at all the points on a given curve pass through a fixed point $(a,b)$. If the curve passes through $(3,-3)$ and $4,-2\sqrt{2}$, and given that $a-2\sqrt{2}b=3$, then $a^2+b^2+ab$ is equal to

Q. The equation of the normal to the curve $y=\sin \frac{\pi x}{2}$ at $(1,1)$ is

1. y = 1
2. x = 1
3. y = x
4. $y-1=\frac{-1}{\pi }(x-1)$

Q. The point of the curve $y^2=2(x-3)$ at which the normal is parallel to the line y - 2x + 1 = 0 is
[MP PET 1998]

1. (5, 2)
2. $(-\frac{1}{2},-2)$
3. (5, -2)
4. $(\frac{3}{2},2)$

Q.

Equation of normal to the curve at y = $4x^2$ at (a, b) is x+16y-258 = 0 . Find the value of (a+b)

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Q. The abscissa of the point on the curve $yx=(x+a{)}^2$ at which the normal cuts off numerically equal intercepts on the coordinate axes is

1. $\frac{2a}{\sqrt{2}}$
2. $\frac{a}{\sqrt{2}}$
3. $2\sqrt{a}$
4. $a$

Q. Let the normals at all the points on a given curve pass through a fixed point $(a,b)$. If the curve passes through $(3,-3)$ and $4,-2\sqrt{2}$, and given that $a-2\sqrt{2}b=3$, then $a^2+b^2+ab$ is equal to

Q. If $x=t^2$and y = 2t, then equation of the normal at t = 1 is
[RPET 1996]

1. x + y - 3 = 0
2. x + y - 1 = 0
3. x + y + 1 = 0
4. x + y + 3 = 0