1-prove-that-the-equaion-of-parabola-whose-axis-of-symmetry-is-parallel-to-y-axisis-given-as-x-h-2-4p-y-k-2-if-Dis-a-perpendicular-distance-of-the-point-p-r-t-from-the-line-L-

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Question Number 182514 by Best1 last updated on 11/Dec/22

$$1.provethattheequaionofparabola$$$$whoseaxisofsymmetryisparallel$$$$toyaxisisgivenas{(x-h)}^2=\underset{-}{+}4p(y-k)$$$$2.ifDisaperpendiculardistance$$$$ofthepointp(r,t)fromtheline$$$$\left(L\right):\frac{x}{r}+\frac{y}{t}=1thenfindthe$$$$valueofD$$$$whynothelpedme??$$

Commented by mr W last updated on 11/Dec/22

$$doyoumeanpeoplemusthelpyou,$$$$nomatteriftheycanor{can}^{\prime }tandif$$$$theywillor{won}^{\prime }t?andtheyshould$$$$tellwhyifthey{don}^{\prime }t\left({can}^{\prime }tor{won}^{\prime }t\right)$$$$helpyou?$$

Answered by mr W last updated on 11/Dec/22

$$2)$$$$distancefrompoint(u,v)toline$$$$ax+bx+c=0isd=\frac{\mid au+bv+c\mid }{\sqrt{a^2+b^2}}.$$$$applyingthisyoucanget$$$$D=\frac{\mid \frac{r}{r}+\frac{t}{t}-1\mid }{\sqrt{\frac{1}{r^2}+\frac{1}{t^2}}}=\frac{\mid rt\mid }{\sqrt{r^2+t^2}}$$

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