differences between object-oriented and component-based design?
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Answer:
\color{magenta}\large\qquad \qquad \underline{ \pmb{{ \mathbb{ \maltese \: \: KEY POINT \: \: \maltese}}}}✠KEYPOINT✠✠KEYPOINT✠
\begin{gathered} \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} \end{gathered}Theequationoftangenthavingslopemtoparabolay2=4axisy=mx+ma
\color{blue} \large\qquad \qquad \underline{ \pmb{{ \mathbb{ \maltese \: \: SOLUTION \: \: \maltese}}}} ✠SOLUTION✠✠SOLUTION✠
\begin{gathered} \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)\end{gathered}Giveneqnofthecurvesarey2=16x(Parabola)...(i)xy=−4(RectangualarHyperbola)...(ii)
\begin{gathered} \mathfrak{Eq {}^{n} \: \: of \: \: tangent \: \: having \: \: slope \: \: \bf {m} } \\ \mathfrak{to \: \: given \: \: parabola \: \: is} \\ \\ \bf y = mx + \frac{4}{m} \: \: \: ...(iii)\end{gathered}Eqnoftangenthavingslopemtogivenparabolaisy=mx+m4...(iii)
Now eliminating y from equations (ii) and (iii)
\large\mathcal{We \: \: Get}WeGet
\begin{gathered}x (mx + \frac{4}{m} ) = - 4 \\ \\ \implies \bf \boxed{ \boxed{{{mx}^{2} + \frac{4}{m} x + 4 = 0}}} \bf \: \: \: ...(iv)\end{gathered}x(mx+m4)=−4⟹mx2+m4x+4=0...(iv)
It will give the points of intersection of tangent and rectangular hyperbola.
\huge \mathcal{As}As
Line (iii) is also a tangent to the rectangular hyperbola.
And there will be only one point of intersection.
\begin{gathered} \therefore \mathfrak{ \text{d}iscriminant \: \: of \: \: the \: \: quadratic \: \: {eq}^{n} \: \: \bf{(iv)}} \\ \large\mathfrak{should \: \: be \: \: zero}\end{gathered}∴discriminantofthequadraticeqn(iv)shouldbezero
\begin{gathered} \implies\sf D = {( \frac{4}{m}) }^{2} - 4(m)(4) = 0 \\ \\ \implies \sf m {}^{3} = 1 \\ \\ \implies \red{ \boxed{ \boxed{ \bf m = 1}}}\end{gathered}⟹D=(m4)2−4(m)(4)=0⟹m3=1⟹m=1
\begin{gathered}\large\mathfrak{So, \: Eq^n \: \: of \: \: reqrd. \: \: tangent \: \: is } \\ \huge \bf y = x + 4\end{gathered}So,Eqnofreqrd.tangentisy=x+4
Refer the attachment
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