Question Number 5614 by Rasheed Soomro last updated on 22/May/16
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{cordinates}\:\mathrm{of}\:\mathrm{the}\:\mathrm{two}\:\mathrm{points} \\ $$$$\mathrm{on}\:\mathrm{the}\:\mathrm{curve}\:\mathrm{y}=\mathrm{4}−\mathrm{x}^{\mathrm{2}} \:\:\mathrm{whose}\:\mathrm{tangents} \\ $$$$\mathrm{pass}\:\mathrm{through}\:\mathrm{the}\:\mathrm{point}\:\left(−\mathrm{1},\:\mathrm{7}\right)\:. \\ $$
Commented by Rasheed Soomro last updated on 24/May/16
$$\mathcal{T}{h}^{\boldsymbol{\alpha}} {nk}^{{S}} ! \\ $$
Commented by Yozzii last updated on 23/May/16
$${Let}\:{the}\:{tangent}\:{to}\:{the}\:{curve}\:{be}\:{given} \\ $$$${by}\:{y}−{y}_{\mathrm{0}} =\frac{{dy}}{{dx}}\mid_{{x}={m}} \left({x}−{x}_{\mathrm{0}} \right)\:{where}\: \\ $$$$\left({x}_{\mathrm{0}} ,{y}_{\mathrm{0}} \right)=\left(−\mathrm{1},\mathrm{7}\right),\:\:\frac{{dy}}{{dx}}=\frac{{d}}{{dx}}\left(\mathrm{4}−{x}^{\mathrm{2}} \right)=−\mathrm{2}{x} \\ $$$${and}\:{x}={m}\:{corresponds}\:{to}\:{a}\:{point} \\ $$$${of}\:{tangency}\:{with}\:{x}={m}. \\ $$$$\therefore\:{y}−\mathrm{7}=\left(−\mathrm{2}{m}\right)\left({x}+\mathrm{1}\right)\: \\ $$$${At}\:{the}\:{point}\:{of}\:{tangency},\:{y}=\mathrm{4}−{m}^{\mathrm{2}} . \\ $$$$\therefore\:\mathrm{4}−{m}^{\mathrm{2}} −\mathrm{7}=−\mathrm{2}{m}\left({m}+\mathrm{1}\right) \\ $$$$−\mathrm{3}−{m}^{\mathrm{2}} =−\mathrm{2}{m}^{\mathrm{2}} −\mathrm{2}{m} \\ $$$${m}^{\mathrm{2}} +\mathrm{2}{m}−\mathrm{3}=\mathrm{0} \\ $$$$\left({m}+\mathrm{3}\right)\left({m}−\mathrm{1}\right)=\mathrm{0} \\ $$$${m}=\mathrm{1},−\mathrm{3}\Rightarrow{y}=\mathrm{3},−\mathrm{5}. \\ $$$$\therefore\:{the}\:{required}\:{points}\:{are}\:\left(\mathrm{1},\mathrm{3}\right),\left(−\mathrm{3},−\mathrm{5}\right). \\ $$$$ \\ $$