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(D) Since the normals at all points on the curve pass through a fixed point $(a, b)$,the curve must be a circle with center $(a, b)$.Let the points on

Vedclass · Accepted Answer

$9$

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(D) Since the normals at all points on the curve pass through a fixed point $(a, b)$,the curve must be a circle with center $(a, b)$.Let the points on the circle be $A(3, -3)$ and $B(4, -2\sqrt{2})$.Since $A$ and $B$ lie on the circle,their distances from the center $C(a, b)$ are equal to the radius $r$.Thus,$CA^{2} = CB^{2}$.$(a - 3)^{2} + (b - (-3))^{2} = (a - 4)^{2} + (b - (-2\sqrt{2}))^{2}$$(a - 3)^{2} + (b + 3)^{2} = (a - 4)^{2} + (b + 2\sqrt{2})^{2}$$a^{2} - 6a + 9 + b^{2} + 6b + 9 = a^{2} - 8a + 16 + b^{2} + 4\sqrt{2}b + 8$$-6a + 6b + 18 = -8a + 4\sqrt{2}b + 24$$2a + (6 - 4\sqrt{2})b = 6$Dividing by $2$,we get $a + (3 - 2\sqrt{2})b = 3$.$a + 3b - 2\sqrt{2}b = 3$$a - 2\sqrt{2}b + 3b = 3 \quad ... (1)$Given that $a - 2\sqrt{2}b = 3 \quad ... (2)$Substituting $(2)$ into $(1)$,we get $3 + 3b = 3$,which implies $3b = 0$,so $b = 0$.Substituting $b = 0$ into $(2)$,we get $a - 2\sqrt{2}(0) = 3$,so $a = 3$.Therefore,$a^{2} + b^{2} + ab = (3)^{2} + (0)^{2} + (3)(0) = 9 + 0 + 0 = 9$.

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 ..... .

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