KCET · Maths · Parabola
For the parabola \(\mathrm{y}^{2}=4 \mathrm{x}\), the point \(P\) whose focal distance is 17 , is
- A \((8,8)\) or \((8,-8)\)
- B \((4,8)\) or \((4,-8)\)
- C \((2,8)\) or \((2,-8)\)
- D \((16,8)\) or \((16,-8)\)
Answer & Solution
Correct Answer
(D) \((16,8)\) or \((16,-8)\)
Step-by-step Solution
Detailed explanation
Given, \(\mathrm{y}^{2}=4 \mathrm{x}\)
Let \(\mathrm{P}(\mathrm{h}, \mathrm{k})\) be any point on the parabola
\(\therefore \quad(\mathrm{h}-1)^{2}+(\mathrm{k}-0)^{2}=17^{2}\)
Also, \(\mathrm{k}^{2}=4 \mathrm{~h}\)
\(\therefore \quad \mathrm{h}^{2}+1-2 \mathrm{~h}+4 \mathrm{~h}=289\)
\(\Rightarrow \quad \mathrm{h}^{2}+2 \mathrm{~h}-288=0\)
\(\Rightarrow \quad(\mathrm{h}+18)(\mathrm{h}-16)=0\)
\(\Rightarrow \quad \mathrm{h}=16 \quad(\because h\) cannot be negative \()\)
\(\therefore \quad \mathrm{k}^{2}=64\)
\(\Rightarrow \quad \mathrm{k}=\pm 8\)
\(\therefore\) Points are \((16,8)\) or \((16,-8)\).
Let \(\mathrm{P}(\mathrm{h}, \mathrm{k})\) be any point on the parabola
\(\therefore \quad(\mathrm{h}-1)^{2}+(\mathrm{k}-0)^{2}=17^{2}\)
Also, \(\mathrm{k}^{2}=4 \mathrm{~h}\)
\(\therefore \quad \mathrm{h}^{2}+1-2 \mathrm{~h}+4 \mathrm{~h}=289\)
\(\Rightarrow \quad \mathrm{h}^{2}+2 \mathrm{~h}-288=0\)
\(\Rightarrow \quad(\mathrm{h}+18)(\mathrm{h}-16)=0\)
\(\Rightarrow \quad \mathrm{h}=16 \quad(\because h\) cannot be negative \()\)
\(\therefore \quad \mathrm{k}^{2}=64\)
\(\Rightarrow \quad \mathrm{k}=\pm 8\)
\(\therefore\) Points are \((16,8)\) or \((16,-8)\).
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