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Question-57958




Question Number 57958 by rahul 19 last updated on 15/Apr/19
Answered by tanmay.chaudhury50@gmail.com last updated on 15/Apr/19
(1,4)  and(3,8)  chord eqn  y−4=(((8−4)/(3−1)))(x−1)  y−4=2x−2  y=2x+2  eqn tangent ∥ to chord is  y=2x+c  solve y=2x+c snd y=x^2 −2x+5  2x+c=x^2 −2x+5  x^2 −4x+5−c=0  B^2 −4AC=0  (−4)^2 −4×1×(5−c)=0  16−20+4c=0  4c=4  c=1  eqn tangent y=2x+1
$$\left(\mathrm{1},\mathrm{4}\right)\:\:{and}\left(\mathrm{3},\mathrm{8}\right) \\ $$$${chord}\:{eqn}\:\:{y}−\mathrm{4}=\left(\frac{\mathrm{8}−\mathrm{4}}{\mathrm{3}−\mathrm{1}}\right)\left({x}−\mathrm{1}\right) \\ $$$${y}−\mathrm{4}=\mathrm{2}{x}−\mathrm{2} \\ $$$${y}=\mathrm{2}{x}+\mathrm{2} \\ $$$${eqn}\:{tangent}\:\parallel\:{to}\:{chord}\:{is} \\ $$$${y}=\mathrm{2}{x}+{c} \\ $$$${solve}\:{y}=\mathrm{2}{x}+{c}\:{snd}\:{y}={x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{5} \\ $$$$\mathrm{2}{x}+{c}={x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{5} \\ $$$${x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{5}−{c}=\mathrm{0} \\ $$$${B}^{\mathrm{2}} −\mathrm{4}{AC}=\mathrm{0} \\ $$$$\left(−\mathrm{4}\right)^{\mathrm{2}} −\mathrm{4}×\mathrm{1}×\left(\mathrm{5}−{c}\right)=\mathrm{0} \\ $$$$\mathrm{16}−\mathrm{20}+\mathrm{4}{c}=\mathrm{0} \\ $$$$\mathrm{4}{c}=\mathrm{4} \\ $$$${c}=\mathrm{1} \\ $$$${eqn}\:{tangent}\:{y}=\mathrm{2}{x}+\mathrm{1} \\ $$$$ \\ $$
Commented by rahul 19 last updated on 15/Apr/19
thank you sir.
$${thank}\:{you}\:{sir}. \\ $$

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