Question

The common tangent of curves x² = 4y and y² = 4x also touches the curve x² + y² = c². then find value of c²?

Answer 1

ANSWER.

=> value of c² = ½

EXPLANATION.

$\sf \to \: common \: tangent \: to \: the \: curve \: {x}^{2} = 4y \: \: and \: \: {y}^{2} = 4x \\ \\ \sf \to \: touches \: the \: curve \: = {x}^{2} + {y}^{2} = {c}^{2}$

To find the value of c².

$\sf \to \: {y}^{2} = 4x \\ \\ \sf \to \: equation \: of \: tangent \: = y = mx + \dfrac{a}{m} \\ \\ \sf \to \: {x}^{2} = 4y \\ \\ \sf \to \: equation \: of \: tangent \: = y = mx - a {m}^{2}$

$\sf \to \: {y}^{2} = 4x \: \implies \: y = mx + \dfrac{1}{m} \: \: ....(1) \\ \\ \sf \to \: x {}^{2} = 4y \implies \: y \: = mx - {m}^{2} .....(2)$

$\sf \to \: compare \: equation \: (1) \: \: and \: \: (2) \: \: we \: get \\ \\ \sf \to \: \dfrac{1}{m} = - {m}^{2} \\ \\ \sf \to \: {m}^{3} = - 1 \\ \\ \sf \to \: m \: = - 1$

$\sf \to \: equation \: of \: tangent \: = y = ( - 1)x - 1 \\ \\ \sf \to \: x \: + y + 1 = 0 \\ \\ \sf \to \: \: it \: lies \: on \: the \: point \: (0 \: 0) \\ \\ \sf \to \: | \frac{1}{ \sqrt{1 + 1} } | = c \\ \\ \sf \to \: {c}^{2} = \dfrac{1}{2}$