Question

Sample Paper 2020 Maths Question 14
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Answer 1

$Given \: \: Question \: \: Is \\ \\ equation \: \: of \: \: curve \: \: is \: \: \: y {}^{2} + 3x - 7 = 0 \\ equation \: of \: line \: parallel \: to \: the \: tangent \:is \\ x - y = 4 \\ find \: \: k \\ \\ \\ Answer \: \\ \\ the \: point \: \: (h \: \: \: k) \: \: lies \: on \: the \: curve \: so \: it \\ must \: satisfies \: the \: equation \: of \: curve. \: \: i.e \\ \\ h {}^{2} + 3k - 7 = 0 \: \: \: ...equation \: \: 01 \\ \\ equation \: of \: parallel \: line \: is \: \: \\ x - y = 4 \: \: \: \: ...equation \: \: 02 \\ \\ slope \: \: of \: \: equation \: \: \: 01 \: \: is \\ \\ 2h \frac{dh}{dk} + 3 = 0 \\ \\ \frac{dh}{dk} = \frac{ - 3}{2h} \\ \\ and \\ \\ slope \: of \: equation \: \: 02 \: \: is \\ \\ 1 - \frac{dy}{dx} = 0 \\ \\ \frac{dy}{dx} = 1 \\ \\ now \: \: both \: the \: lines \: are \: parallel \: so \: their \: slopes \\ are \: equal \\ \\ 1 = \frac{ - 3}{2h} \\ \\ 2h = - 3 \\ \\ h = \frac{ - 3}{2} \\ \\ now \: \: substitute \: \: the \: value \: of \: h \: \\ in \: equation \: \: 01 \\ \\ ( \frac{ - 3}{2} ) {}^{2} + 3k - 7 = 0 \\ \\ \frac{9}{4} - 7 = - 3k \\ \\ - 3k = \frac{ - 19}{4} \\ \\ k = \frac{19}{12} \\ \\ therefore \: \: \: k = \frac{19}{12}$