Question

#Quality Question

@Coordinate Geometry

Q》》
The equation of the common tangent to the Curves,
${y}^{2} = 16x \: \: and \: \: xy = - 4 \: \: \: \bf \: is$
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Answer 1

$\color{magenta}\large\qquad \qquad \underline{ \pmb{{ \mathbb{ \maltese \: \: KEY POINT \: \: \maltese}}}}$

$\mathfrak{ \text{T}he \: \: equation \: \: of \: \: tangent \: \: having \: \: slope \: \: \bf{ m} } \\ \\ \mathfrak{to \: \: parabola \: \: { \bf{y}^{2} = 4ax } \: \: is } \\ \\ \bf y = mx + \frac{a}{m}$

$\color{blue} \large\qquad \qquad \underline{ \pmb{{ \mathbb{ \maltese \: \: SOLUTION \: \: \maltese}}}}$

$\mathfrak{Given \: \: eq {}^{n} \: \: of \: \: the \: \: curves \: \: are} \\ \\ \bf {y}^{2} = 16x \: \: ( \mathfrak{ Parabola}) \: \: \: ...(i) \\ \\ \bf xy = - 4( \mathfrak{Rectangualar \: H \text{y}perbola)}...(ii)$

$\mathfrak{Eq {}^{n} \: \: of \: \: tangent \: \: having \: \: slope \: \: \bf {m} } \\ \mathfrak{to \: \: given \: \: parabola \: \: is} \\ \\ \bf y = mx + \frac{4}{m} \: \: \: ...(iii)$

Now eliminating y from equations (ii) and (iii)

$\large\mathcal{We \: \: Get}$

$x (mx + \frac{4}{m} ) = - 4 \\ \\ \implies \bf \boxed{ \boxed{{{mx}^{2} + \frac{4}{m} x + 4 = 0}}} \bf \: \: \: ...(iv)$

It will give the points of intersection of tangent and rectangular hyperbola.

$\huge \mathcal{As}$

Line (iii) is also a tangent to the rectangular hyperbola.

And there will be only one point of intersection.

$\therefore \mathfrak{ \text{d}iscriminant \: \: of \: \: the \: \: quadratic \: \: {eq}^{n} \: \: \bf{(iv)}} \\ \large\mathfrak{should \: \: be \: \: zero}$

$\implies\sf D = {( \frac{4}{m}) }^{2} - 4(m)(4) = 0 \\ \\ \implies \sf m {}^{3} = 1 \\ \\ \implies \red{ \boxed{ \boxed{ \bf m = 1}}}$

$\large\mathfrak{So, \: Eq^n \: \: of \: \: reqrd. \: \: tangent \: \: is } \\ \huge \bf y = x + 4$

Answer 2

Solution

$\mathcal{Given \: \: eq {}^{n}s \: \: are} \\ \\ \bf {y}^{2} = 16x \: \: \: \: \: ...(i) \\ \\ \bf xy = - 4...(ii)$

$\mathcal{Eq {}^{n} \: \: of \: \: tangent \: \: having \: \: slope \: \: ' {m} ' } \\ \mathcal{to \: \: given \: \: parabola(i) \: \: is} \\ \\ \bf y = mx + \frac{4}{m} \: \: \: ...(iii)$

Solving equation (ii) and (iii)

$\large\mathfrak{We \: \: Get}$

$x (mx + \frac{4}{m} ) = - 4 \\ \\ \implies {mx}^{2} + \frac{4}{m} x + 4 = 0\: \: \: ...(iv)$

For required condition:

${ \text{d}iscriminant \: \: of \: \: the \: \: {eq}^{n} \: \: \bf{(iv)}} \\ \large\mathcal{Will \: \: be \: \: zero}$

$D = {( \frac{4}{m}) }^{2} - 4(m)(4) = 0 \\ \\ \implies \sf m {}^{3} = 1 \\ \\ \implies \blue{ \boxed{ \bf m = 1}}$

$\large\mathcal{So, \: Eq^n \: \: of \: \: tangent \: \: is } \\ \ \bf y = x + 4$