+10점 · 접선의 방정식을 올바르게 세웠는가 (a) 포물선 y 2 = 4 x \displaystyle y^{2}=4x y 2 = 4 x ( t 2 , 2 t ) \displaystyle (t^{2},\,2t) ( t 2 , 2 t ) ( 2 t ) y = 2 ( x + t 2 ) \displaystyle (2t)y=2\left(x+t^{2}\right) ( 2 t ) y = 2 ( x + t 2 ) t y = x + t 2 \displaystyle ty=x+t^{2} t y = x + t 2
Tip. 포물선 y 2 = 4 x \displaystyle y^{2}=4x y 2 = 4 x ( t 2 , 2 t ) \displaystyle (t^{2},\,2t) ( t 2 , 2 t )
+10점 · 접선이 P ( α , β ) \displaystyle \mathrm{P}(\alpha,\,\beta) P ( α , β ) 이 접선이 P ( α , β ) \displaystyle \mathrm{P}(\alpha,\,\beta) P ( α , β ) t β = α + t 2 \displaystyle t\beta=\alpha+t^{2} tβ = α + t 2 t 2 − β t + α = 0 ⋯ ( ∗ ) \displaystyle t^{2}-\beta t+\alpha=0\ \cdots\ (\ast) t 2 − β t + α = 0 ⋯ ( ∗ )
+10점 · 두 접점이 그 방정식의 서로 다른 두 실근에 대응함을 언급했는가 두 접점 Q ( t 1 2 , 2 t 1 ) \displaystyle \mathrm{Q}(t_{1}^{2},\,2t_{1}) Q ( t 1 2 , 2 t 1 ) R ( t 2 2 , 2 t 2 ) \displaystyle \mathrm{R}(t_{2}^{2},\,2t_{2}) R ( t 2 2 , 2 t 2 ) t 1 \displaystyle t_{1} t 1 t 2 \displaystyle t_{2} t 2 ( ∗ ) \displaystyle (\ast) ( ∗ )
+5점 · 판별식 조건 β 2 − 4 α > 0 \displaystyle \beta^{2}-4\alpha>0 β 2 − 4 α > 0 따라서 판별식에서 β 2 − 4 α > 0 \displaystyle \beta^{2}-4\alpha>0 β 2 − 4 α > 0
+5점 · 근과 계수와의 관계를 활용했는가 근과 계수와의 관계에서 t 1 + t 2 = β \displaystyle t_{1}+t_{2}=\beta t 1 + t 2 = β t 1 t 2 = α \displaystyle t_{1}t_{2}=\alpha t 1 t 2 = α
Caution. 서로 다른 실근의 존재 조건을 확인해야 한다 : 이 문제에서는 서로 다른 두 접점이 존재해야 하므로, 판별식 β 2 − 4 α > 0 \displaystyle \beta^2-4\alpha>0 β 2 − 4 α > 0
+10점 · P Q → ⋅ P R → \displaystyle \overrightarrow{\mathrm{PQ}}\cdot\overrightarrow{\mathrm{PR}} PQ ⋅ PR P Q → = ( t 1 2 − α , 2 t 1 − β ) \displaystyle \overrightarrow{\mathrm{PQ}}=\left(t_{1}^{2}-\alpha,\,2t_{1}-\beta\right) PQ = ( t 1 2 − α , 2 t 1 − β ) P R → = ( t 2 2 − α , 2 t 2 − β ) \displaystyle \overrightarrow{\mathrm{PR}}=\left(t_{2}^{2}-\alpha,\,2t_{2}-\beta\right) PR = ( t 2 2 − α , 2 t 2 − β ) P Q → ⋅ P R → = ( t 1 2 − α ) ( t 2 2 − α ) + ( 2 t 1 − β ) ( 2 t 2 − β ) \displaystyle \overrightarrow{\mathrm{PQ}}\cdot\overrightarrow{\mathrm{PR}}=\left(t_{1}^{2}-\alpha\right)\left(t_{2}^{2}-\alpha\right)+\left(2t_{1}-\beta\right)\left(2t_{2}-\beta\right) PQ ⋅ PR = ( t 1 2 − α ) ( t 2 2 − α ) + ( 2 t 1 − β ) ( 2 t 2 − β )
+10점 · 대칭식 계산을 올바르게 수행했는가 이때 t 1 2 + t 2 2 = ( t 1 + t 2 ) 2 − 2 t 1 t 2 = β 2 − 2 α \displaystyle t_{1}^{2}+t_{2}^{2}=\left(t_{1}+t_{2}\right)^{2}-2t_{1}t_{2}=\beta^{2}-2\alpha t 1 2 + t 2 2 = ( t 1 + t 2 ) 2 − 2 t 1 t 2 = β 2 − 2 α ( t 1 2 − α ) ( t 2 2 − α ) = ( t 1 t 2 ) 2 − α ( t 1 2 + t 2 2 ) + α 2 = α 2 − α ( β 2 − 2 α ) + α 2 = α ( 4 α − β 2 ) \displaystyle \left(t_{1}^{2}-\alpha\right)\left(t_{2}^{2}-\alpha\right)=\left(t_{1}t_{2}\right)^{2}-\alpha\left(t_{1}^{2}+t_{2}^{2}\right)+\alpha^{2}=\alpha^{2}-\alpha\left(\beta^{2}-2\alpha\right)+\alpha^{2}=\alpha\left(4\alpha-\beta^{2}\right) ( t 1 2 − α ) ( t 2 2 − α ) = ( t 1 t 2 ) 2 − α ( t 1 2 + t 2 2 ) + α 2 = α 2 − α ( β 2 − 2 α ) + α 2 = α ( 4 α − β 2 ) ( 2 t 1 − β ) ( 2 t 2 − β ) = 4 t 1 t 2 − 2 β ( t 1 + t 2 ) + β 2 = 4 α − β 2 \displaystyle \left(2t_{1}-\beta\right)\left(2t_{2}-\beta\right)=4t_{1}t_{2}-2\beta\left(t_{1}+t_{2}\right)+\beta^{2}=4\alpha-\beta^{2} ( 2 t 1 − β ) ( 2 t 2 − β ) = 4 t 1 t 2 − 2 β ( t 1 + t 2 ) + β 2 = 4 α − β 2
+10점 · P Q → ⋅ P R → = ( α + 1 ) ( 4 α − β 2 ) \displaystyle \overrightarrow{\mathrm{PQ}}\cdot\overrightarrow{\mathrm{PR}}=\left(\alpha+1\right)\left(4\alpha-\beta^{2}\right) PQ ⋅ PR = ( α + 1 ) ( 4 α − β 2 ) ∴ P Q → ⋅ P R → = α ( 4 α − β 2 ) + ( 4 α − β 2 ) = ( α + 1 ) ( 4 α − β 2 ) \displaystyle \therefore\ \overrightarrow{\mathrm{PQ}}\cdot\overrightarrow{\mathrm{PR}}=\alpha\left(4\alpha-\beta^{2}\right)+\left(4\alpha-\beta^{2}\right)=\left(\alpha+1\right)\left(4\alpha-\beta^{2}\right) ∴ PQ ⋅ PR = α ( 4 α − β 2 ) + ( 4 α − β 2 ) = ( α + 1 ) ( 4 α − β 2 )
+5점 · 직각 조건을 P Q → ⋅ P R → = 0 \displaystyle \overrightarrow{\mathrm{PQ}}\cdot\overrightarrow{\mathrm{PR}}=0 PQ ⋅ PR = 0 (b) ∠ Q P R \displaystyle \angle \mathrm{QPR} ∠ QPR P Q → ⋅ P R → = 0 \displaystyle \overrightarrow{\mathrm{PQ}}\cdot\overrightarrow{\mathrm{PR}}=0 PQ ⋅ PR = 0
+8점 · 4 α − β 2 < 0 \displaystyle 4\alpha-\beta^{2}<0 4 α − β 2 < 0 α = − 1 \displaystyle \alpha=-1 α = − 1 (a)에서 P Q → ⋅ P R → = ( α + 1 ) ( 4 α − β 2 ) \displaystyle \overrightarrow{\mathrm{PQ}}\cdot\overrightarrow{\mathrm{PR}}=\left(\alpha+1\right)\left(4\alpha-\beta^{2}\right) PQ ⋅ PR = ( α + 1 ) ( 4 α − β 2 ) β 2 − 4 α > 0 \displaystyle \beta^{2}-4\alpha>0 β 2 − 4 α > 0 4 α − β 2 < 0 \displaystyle 4\alpha-\beta^{2}<0 4 α − β 2 < 0 P Q → ⋅ P R → = 0 \displaystyle \overrightarrow{\mathrm{PQ}}\cdot\overrightarrow{\mathrm{PR}}=0 PQ ⋅ PR = 0 α = − 1 \displaystyle \alpha=-1 α = − 1
+7점 · α = − 1 \displaystyle \alpha=-1 α = − 1 β \displaystyle \beta β β 2 + 4 > 0 \displaystyle \beta^{2}+4>0 β 2 + 4 > 0 이때 β 2 − 4 α = β 2 + 4 > 0 \displaystyle \beta^{2}-4\alpha=\beta^{2}+4>0 β 2 − 4 α = β 2 + 4 > 0 β \displaystyle \beta β
+5점 · 준선 x = − 1 \displaystyle x=-1 x = − 1 ∴ P = ( − 1 , β ) \displaystyle \therefore\ \mathrm{P}=(-1,\,\beta) ∴ P = ( − 1 , β ) β \displaystyle \beta β x = − 1 \displaystyle x=-1 x = − 1
수리논술, 이게 다다.